Search: seq:1,0,1,0,1,1,0,2,3,1
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A264428
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Triangle read by rows, Bell transform of Bell numbers.
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+30
187
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1, 0, 1, 0, 1, 1, 0, 2, 3, 1, 0, 5, 11, 6, 1, 0, 15, 45, 35, 10, 1, 0, 52, 205, 210, 85, 15, 1, 0, 203, 1029, 1330, 700, 175, 21, 1, 0, 877, 5635, 8946, 5845, 1890, 322, 28, 1, 0, 4140, 33387, 63917, 50358, 20055, 4410, 546, 36, 1, 0, 21147, 212535, 484140, 450905, 214515, 57855, 9240, 870, 45, 1
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OFFSET
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0,8
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COMMENTS
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Consider the sequence S0 -> T0 -> S1 -> T1 -> S2 -> T2 -> ... Here Sn -> Tn indicates the Bell transform mapping a sequence Sn to a triangle Tn as defined in the link and Tn -> S{n+1} the operator associating a triangle with the sequence of its row sums. If
T0 = A048993 # Stirling subset numbers,
T1 = A264428 # Bell transform of Bell numbers,
S2 = A187761 # second-order Bell numbers,
T2 = A264430 # Bell transform of second-order Bell numbers,
S3 = A264432 # third-order Bell numbers.
This construction is closely related to permutations trees and A179455. Sn is A179455_col(n+1) prepended by A179455_diag(k) = k! for k <= n. In other words, Sn 'converges' to n! for n -> oo.
Given a sequence (s(n))n>=0 with s(0) = 0 and with e.g.f. B(x) = Sum_{n >= 1} s(n)*x^n/n!, then the Bell matrix associated with s(n) equals the exponential Riordan array [1, B(x)] belonging to the Lagrange subgroup of the exponential Riordan group. Omitting the first row and column from the Bell matrix produces the exponential Riordan array [d/dx(B(x)), B(x)] belonging to the Derivative subgroup of the exponential Riordan group. - Peter Bala, Jun 07 2016
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LINKS
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FORMULA
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E.g.f.: exp(t*B(x)), where B(x) = Integral_{u = 0..x} exp(exp(u) - 1) du = x + x^2/2! + 2*x^3/3! + 5*x^4/4! + 15*x^5/5! + 52*x^6/6! + ....
Row polynomial recurrence: R(n+1,t) = t*Sum_{k = 0 ..n} binomial(n,k)*Bell(k)* R(n-k,t) with R(0,t) = 1. (End)
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EXAMPLE
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Triangle starts:
[1]
[0, 1]
[0, 1, 1]
[0, 2, 3, 1]
[0, 5, 11, 6, 1]
[0, 15, 45, 35, 10, 1]
[0, 52, 205, 210, 85, 15, 1]
[0, 203, 1029, 1330, 700, 175, 21, 1]
[0, 877, 5635, 8946, 5845, 1890, 322, 28, 1]
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MAPLE
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# Computes sequence in matrix form.
BellMatrix := proc(f, len) local T, A; A := [seq(f(n), n=0..len-2)];
T := proc(n, k) option remember; if k=0 then k^n else
add(binomial(n-1, j-1)*T(n-j, k-1)*A[j], j=1..n-k+1) fi end;
Matrix(len, (n, k)->T(n-1, k-1), shape=triangular[lower]) end:
BellMatrix(n -> combinat:-bell(n), 9); # Peter Luschny, Jan 21 2016
# Alternative, using the recurrence of Peter Bala:
R := proc(n) option remember; if n = 0 then 1 else
t*add(binomial(n-1, k)*combinat:-bell(k)*R(n-k-1, t), k=0..n-1) fi end:
T_row := n-> seq(coeff(R(n), t, k), k=0..n):
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MATHEMATICA
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BellMatrix[f_Function|f_Symbol, len_] := With[{t = Array[f, len, 0]}, Table[BellY[n, k, t], {n, 0, len-1}, {k, 0, len-1}]];
rows = 11;
M = BellMatrix[BellB, rows];
Table[M[[n, k]], {n, 1, rows}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jan 21 2016, updated Jul 14 2018 *)
With[{r = 8}, Flatten[Table[BellY[n, k, BellB[Range[0, r]]], {n, 0, r}, {k, 0, n}]]] (* Jan Mangaldan, May 22 2016 *)
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PROG
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(Sage)
# The functions below are referenced in various other sequences.
def bell_transform(n, a): # partition_based
row = []
fn = factorial(n)
for k in (0..n):
result = 0
for p in Partitions(n, length=k):
factorial_product = 1
power_factorial_product = 1
for part, count in p.to_exp_dict().items():
factorial_product *= factorial(count)
power_factorial_product *= factorial(part)**count
coefficient = fn//(factorial_product*power_factorial_product)
result += coefficient*prod([a[i-1] for i in p])
row.append(result)
return row
def bell_matrix(generator, dim):
G = [generator(k) for k in srange(dim)]
row = lambda n: bell_transform(n, G)
return matrix(ZZ, [row(n)+[0]*(dim-n-1) for n in srange(dim)])
def inverse_bell_matrix(generator, dim):
G = [generator(k) for k in srange(dim)]
row = lambda n: bell_transform(n, G)
M = matrix(ZZ, [row(n)+[0]*(dim-n-1) for n in srange(dim)]).inverse()
return matrix(ZZ, dim, lambda n, k: (-1)^(n-k)*M[n, k])
bell_numbers = [sum(bell_transform(n, [1]*10)) for n in range(11)]
for n in range(11): print(bell_transform(n, bell_numbers))
(PARI)
bell_matrix(f, len) = { my( m = matrix(len, len) ); m[1, 1] = 1;
for( n = 1, len-1, m[n+1, 2] = f(n-1) );
for( n = 0, len-1, for( k = 1, n,
m[n+1, k+1] = sum(j = 1, n-k+1, binomial(n-1, j-1)*m[n-j+1, k]*m[j+1, 2]) ));
return( m )
}
f(n) = polcoeff( sum( k=0, n, prod( i=1, k, x / (1 - i*x)), x^n * O(x)), n);
(Python)
from functools import cache
from math import comb as binomial
def BellMatrix(f, size):
A = [f(n) for n in range(size - 1)]
@cache
def T(n, k):
if k == 0: return k ** n
return sum(
binomial(n - 1, j) * T(n - j - 1, k - 1) * A[j]
for j in range(n - k + 1) )
return [[T(n, k) for k in range(n + 1)] for n in range(size)]
@cache
def b(n, k=0): return n < 1 or k*b(n-1, k) + b(n-1, k+1)
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CROSSREFS
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Cf. A000012, A000110, A000217, A000914, A027801, A048993, A051836, A179455, A187761 (row sums), A264430, A264432, A265312.
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KEYWORD
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AUTHOR
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STATUS
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approved
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A132393
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Triangle of unsigned Stirling numbers of the first kind (see A048994), read by rows, T(n,k) for 0 <= k <= n.
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+30
115
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1, 0, 1, 0, 1, 1, 0, 2, 3, 1, 0, 6, 11, 6, 1, 0, 24, 50, 35, 10, 1, 0, 120, 274, 225, 85, 15, 1, 0, 720, 1764, 1624, 735, 175, 21, 1, 0, 5040, 13068, 13132, 6769, 1960, 322, 28, 1, 0, 40320, 109584, 118124, 67284, 22449, 4536, 546, 36, 1, 0, 362880, 1026576, 1172700
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OFFSET
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0,8
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COMMENTS
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Another name: Triangle of signless Stirling numbers of the first kind.
Triangle T(n,k), 0<=k<=n, read by rows given by [0,1,1,2,2,3,3,4,4,5,5,...] DELTA [1,0,1,0,1,0,1,0,1,...] where DELTA is the operator defined in A084938.
Exponential Riordan array [1/(1-x), log(1/(1-x))]. - Ralf Stephan, Feb 07 2014
Also the Bell transform of the factorial numbers (A000142). For the definition of the Bell transform see A264428 and for cross-references A265606. - Peter Luschny, Dec 31 2015
This is the lower triagonal Sheffer matrix of the associated or Jabotinsky type |S1| = (1, -log(1-x)) (see the W. Lang link under A006232 for the notation and references). This implies the e.g.f.s given below. |S1| is the transition matrix from the monomial basis {x^n} to the rising factorial basis {risefac(x,n)}, n >= 0. - Wolfdieter Lang, Feb 21 2017
T(n, k), for n >= k >= 1, is also the total volume of the n-k dimensional cell (polytope) built from the n-k orthogonal vectors of pairwise different lengths chosen from the set {1, 2, ..., n-1}. See the elementary symmetric function formula for T(n, k) and an example below. - Wolfdieter Lang, May 28 2017
The compositional inverse w.r.t. x of y = y(t;x) = x*(1 - t(-log(1-x)/x)) = x + t*log(1-x) is x = x(t;y) = ED(y,t) := Sum_{d>=0} D(d,t)*y^(d+1)/(d+1)!, the e.g.f. of the o.g.f.s D(d,t) = Sum_{m>=0} T(d+m, m)*t^m of the diagonal sequences of the present triangle. See the P. Bala link for a proof (there d = n-1, n >= 1, is the label for the diagonals).
This inversion gives D(d,t) = P(d, t)/(1-t)^(2*d+1), with the numerator polynomials P(d, t) = Sum_{m=0..d} A288874(d, m)*t^m. See an example below. See also the P. Bala formula in A112007. (End)
For n > 0, T(n,k) is the number of permutations of the integers from 1 to n which have k visible digits when viewed from a specific end, in the sense that a higher value hides a lower one in a subsequent position. - Ian Duff, Jul 12 2019
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REFERENCES
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Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, pages 31, 187, 441, 996.
R. L. Graham, D. E. Knuth, and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 2nd. ed., Table 259, p. 259.
Steve Roman, The Umbral Calculus, Dover Publications, New York (1984), pp. 149-150
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LINKS
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FORMULA
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T(n,k) = T(n-1,k-1)+(n-1)*T(n-1,k), n,k>=1; T(n,0)=T(0,k); T(0,0)=1.
Sum_{k=0..n} T(n,k)*x^(n-k) = A000012(n), A000142(n), A001147(n), A007559(n), A007696(n), A008548(n), A008542(n), A045754(n), A045755(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8 respectively. - Philippe Deléham, Nov 13 2007
Expand 1/(1-t)^x = Sum_{n>=0}p(x,n)*t^n/n!; then the coefficients of the p(x,n) produce the triangle. - Roger L. Bagula, Apr 18 2008
Sum_{k=0..n} T(n,k)*2^k*x^(n-k) = A000142(n+1), A000165(n), A008544(n), A001813(n), A047055(n), A047657(n), A084947(n), A084948(n), A084949(n) for x = 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively. - Philippe Deléham, Sep 18 2008
a(n) = Sum_{k=0..n} T(n,k)*3^k*x^(n-k) = A001710(n+2), A001147(n+1), A032031(n), A008545(n), A047056(n), A011781(n), A144739(n), A144756(n), A144758(n) for x=1,2,3,4,5,6,7,8,9,respectively. - Philippe Deléham, Sep 20 2008
Sum_{k=0..n} T(n,k)*4^k*x^(n-k) = A001715(n+3), A002866(n+1), A007559(n+1), A047053(n), A008546(n), A049308(n), A144827(n), A144828(n), A144829(n) for x=1,2,3,4,5,6,7,8,9 respectively. - Philippe Deléham, Sep 21 2008
Sum_{k=0..n} x^k*T(n,k) = x*(1+x)*(2+x)*...*(n-1+x), n>=1. - Philippe Deléham, Oct 17 2008
E.g.f. k-th column: (-log(1 - x))^k, k >= 0.
E.g.f. triangle (see the Apr 18 2008 Baluga comment): exp(-x*log(1-z)).
E.g.f. a-sequence: x/(1 - exp(-x)). See A164555/A027642. The e.g.f. for the z-sequence is 0. (End)
The row polynomials R(n, x) = Sum_{k=0..n} T(n, k)*x^k, for n >= 0, are R(n, x) = risefac(x,n-1) := Product_{j=0..n-1} x+j, with the empty product for n=0 put to 1. See the Feb 21 2017 comment above. This implies:
T(n, k) = sigma^{(n-1)}_(n-k), for n >= k >= 1, with the elementary symmetric functions sigma^{(n-1))_m of degree m in the n-1 symbols 1, 2, ..., n-1, with binomial(n-1, m) terms. See an example below.(End)
Boas-Buck type recurrence for column sequence k: T(n, k) = (n!*k/(n - k)) * Sum_{p=k..n-1} beta(n-1-p)*T(p, k)/p!, for n > k >= 0, with input T(k, k) = 1, and beta(k) = A002208(k+1)/A002209(k+1). See a comment and references in A286718. - Wolfdieter Lang, Aug 11 2017
T(n,k) = Sum_{j=k..n} j^(j-k)*binomial(j-1, k-1)*A354795(n,j) for n > 0. - Mélika Tebni, Mar 02 2023
n-th row polynomial: n!*Sum_{k = 0..2*n} (-1)^k*binomial(-x, k)*binomial(-x, 2*n-k) = n!*Sum_{k = 0..2*n} (-1)^k*binomial(1-x, k)*binomial(-x, 2*n-k). - Peter Bala, Mar 31 2024
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EXAMPLE
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Triangle T(n,k) begins:
1;
0, 1;
0, 1, 1;
0, 2, 3, 1;
0, 6, 11, 6, 1;
0, 24, 50, 35, 10, 1;
0, 120, 274, 225, 85, 15, 1;
0, 720, 1764, 1624, 735, 175, 21, 1;
0, 5040, 13068, 13132, 6769, 1960, 322, 28, 1;
---------------------------------------------------
Production matrix is
0, 1
0, 1, 1
0, 1, 2, 1
0, 1, 3, 3, 1
0, 1, 4, 6, 4, 1
0, 1, 5, 10, 10, 5, 1
0, 1, 6, 15, 20, 15, 6, 1
0, 1, 7, 21, 35, 35, 21, 7, 1
...
Three term recurrence: 50 = T(5, 2) = 1*6 + (5-1)*11 = 50.
Recurrence from the Sheffer a-sequence [1, 1/2, 1/6, 0, ...]: 50 = T(5, 2) = (5/2)*(binomial(1, 1)*1*6 + binomial(2, 1)*(1/2)*11 + binomial(3, 1)*(1/6)*6 + 0) = 50. The vanishing z-sequence produces the k=0 column from T(0, 0) = 1. (End)
Elementary symmetric function T(4, 2) = sigma^{(3)}_2 = 1*2 + 1*3 + 2*3 = 11. Here the cells (polytopes) are 3 rectangles with total area 11. - Wolfdieter Lang, May 28 2017
O.g.f.s of diagonals: d=2 (third diagonal) [0, 6, 50, ...] has D(2,t) = P(2, t)/(1-t)^5, with P(2, t) = 2 + t, the n = 2 row of A288874. - Wolfdieter Lang, Jul 20 2017
Boas-Buck recurrence for column k = 2 and n= 5: T(5, 2) = (5!*2/3)*((3/8)*T(2,2)/2! + (5/12)*T(3,2)/3! + (1/2)*T(4,2)/4!) = (5!*2/3)*((3/16 + (5/12)*3/3! + (1/2)*11/4!) = 50. The beta sequence begins: {1/2, 5/12, 3/8, ...}. - Wolfdieter Lang, Aug 11 2017
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MAPLE
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a132393_row := proc(n) local k; seq(coeff(expand(pochhammer (x, n)), x, k), k=0..n) end: # Peter Luschny, Nov 28 2010
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MATHEMATICA
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p[t_] = 1/(1 - t)^x; Table[ ExpandAll[(n!)SeriesCoefficient[ Series[p[t], {t, 0, 30}], n]], {n, 0, 10}]; a = Table[(n!)* CoefficientList[SeriesCoefficient[ Series[p[t], {t, 0, 30}], n], x], {n, 0, 10}]; Flatten[a] (* Roger L. Bagula, Apr 18 2008 *)
Flatten[Table[Abs[StirlingS1[n, i]], {n, 0, 10}, {i, 0, n}]] (* Harvey P. Dale, Feb 04 2014 *)
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PROG
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(Maxima) create_list(abs(stirling1(n, k)), n, 0, 12, k, 0, n); /* Emanuele Munarini, Mar 11 2011 */
(Haskell)
a132393 n k = a132393_tabl !! n !! k
a132393_row n = a132393_tabl !! n
a132393_tabl = map (map abs) a048994_tabl
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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A145142
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Triangle T(n,k), n>=1, 0<=k<=n-1, read by rows: T(n,k)/(n-1)! is the coefficient of x^k in polynomial p_n for the n-th row sequence of A145153.
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+30
18
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1, 0, 1, 0, 1, 1, 0, 2, 3, 1, 24, 6, 11, 6, 1, 120, 144, 50, 35, 10, 1, 720, 1200, 634, 225, 85, 15, 1, 5040, 9960, 6804, 2464, 735, 175, 21, 1, 80640, 89040, 71868, 29932, 8449, 1960, 322, 28, 1, 1088640, 1231776, 789984, 375164, 112644, 25473, 4536, 546, 36, 1
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OFFSET
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1,8
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LINKS
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FORMULA
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See program.
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EXAMPLE
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Triangle begins:
1;
0, 1;
0, 1, 1;
0, 2, 3, 1;
24, 6, 11, 6, 1;
120, 144, 50, 35, 10, 1;
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MAPLE
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row:= proc(n) option remember; local f, i, x; f:= unapply(simplify(sum('cat(a||i) *x^i', 'i'=0..n-1) ), x); unapply(subs(solve({seq(f(i+1)= coeftayl(x/ (1-x-x^4)/ (1-x)^i, x=0, n), i=0..n-1)}, {seq(cat(a||i), i=0..n-1)}), sum('cat(a||i) *x^i', 'i'=0..n-1) ), x); end: T:= (n, k)-> `if`(k<0 or k>=n, 0, coeff(row(n)(x), x, k)*(n-1)!): seq(seq(T(n, k), k=0..n-1), n=1..12);
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MATHEMATICA
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row[n_] := Module[{f, eq}, f = Function[x, Sum[a[k]*x^k, {k, 0, n-1}]]; eq = Table[f[k+1] == SeriesCoefficient[x/(1-x-x^4)/(1-x)^k, {x, 0, n}], {k, 0, n-1}]; Table[a[k], {k, 0, n-1}] /. Solve[eq] // First]; Table[row[n]*(n-1)!, {n, 1, 12}] // Flatten (* Jean-François Alcover, Feb 04 2014, after Alois P. Heinz *)
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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A293113
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Number T(n,k) of sets of nonempty words with a total of n letters over k-ary alphabet containing the k-th letter such that within each prefix of a word every letter of the alphabet is at least as frequent as the subsequent alphabet letter; triangle T(n,k), n>=0, 0<=k<=n, read by rows.
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+30
14
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1, 0, 1, 0, 1, 1, 0, 2, 3, 1, 0, 2, 8, 4, 1, 0, 3, 20, 16, 5, 1, 0, 4, 47, 53, 25, 6, 1, 0, 5, 106, 173, 102, 36, 7, 1, 0, 6, 237, 532, 410, 172, 49, 8, 1, 0, 8, 522, 1615, 1545, 813, 268, 64, 9, 1, 0, 10, 1146, 4785, 5784, 3576, 1448, 394, 81, 10, 1
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OFFSET
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0,8
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LINKS
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FORMULA
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EXAMPLE
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Triangle T(n,k) begins:
1;
0, 1;
0, 1, 1;
0, 2, 3, 1;
0, 2, 8, 4, 1;
0, 3, 20, 16, 5, 1;
0, 4, 47, 53, 25, 6, 1;
0, 5, 106, 173, 102, 36, 7, 1;
0, 6, 237, 532, 410, 172, 49, 8, 1;
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MAPLE
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h:= l-> (n-> add(i, i=l)!/mul(mul(1+l[i]-j+add(`if`(l[k]
<j, 0, 1), k=i+1..n), j=1..l[i]), i=1..n))(nops(l)):
g:= proc(n, i, l) option remember;
`if`(n=0, h(l), `if`(i<1, 0, `if`(i=1, h([l[], 1$n]),
g(n, i-1, l) +`if`(i>n, 0, g(n-i, i, [l[], i])))))
end:
b:= proc(n, i, k) option remember; `if`(n=0, 1, `if`(i<1, 0,
add(b(n-i*j, i-1, k)*binomial(g(i, k, []), j), j=0..n/i)))
end:
T:= (n, k)-> b(n$2, k)-`if`(k=0, 0, b(n$2, k-1)):
seq(seq(T(n, k), k=0..n), n=0..14);
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MATHEMATICA
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h[l_] := Function[n, Total[l]!/Product[Product[1 + l[[i]] - j + Sum[If[ l[[k]]<j, 0, 1], {k, i+1, n}], {j, 1, l[[i]]}], {i, 1, n}]][ Length[l]];
g[n_, i_, l_] := g[n, i, l] = If[n == 0, h[l], If[i < 1, 0, If[i == 1, h[Join[l, Table[1, n]]], g[n, i - 1, l] + If[i > n, 0, g[n - i, i, Append[l, i]]]]]];
b[n_, i_, k_] := b[n, i, k] = If[n == 0, 1, If[i<1, 0, Sum[b[n - i*j, i-1, k]*Binomial[g[i, k, {}], j], {j, 0, n/i}]]];
T[n_, k_] := b[n, n, k] - If[k == 0, 0, b[n, n, k - 1]];
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CROSSREFS
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Columns k=0-10 give: A000007, A000009 (for n>0), A293883, A293884, A293885, A293886, A293887, A293888, A293889, A293890, A293891.
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KEYWORD
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AUTHOR
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STATUS
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approved
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A344855
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Number T(n,k) of permutations of [n] having k cycles of the form (c1, c2, ..., c_m) where c1 = min_{i>=1} c_i and c_j = min_{i>=j} c_i or c_j = max_{i>=j} c_i; triangle T(n,k), n>=0, 0<=k<=n, read by rows.
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+30
13
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1, 0, 1, 0, 1, 1, 0, 2, 3, 1, 0, 4, 11, 6, 1, 0, 8, 40, 35, 10, 1, 0, 16, 148, 195, 85, 15, 1, 0, 32, 560, 1078, 665, 175, 21, 1, 0, 64, 2160, 5992, 5033, 1820, 322, 28, 1, 0, 128, 8448, 33632, 37632, 17913, 4284, 546, 36, 1, 0, 256, 33344, 190800, 280760, 171465, 52941, 9030, 870, 45, 1
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OFFSET
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0,8
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COMMENTS
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The sequence of column k satisfies a linear recurrence with constant coefficients of order k*(k+1)/2 = A000217(k).
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LINKS
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FORMULA
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Sum_{k=1..n} k * T(n,k) = A345341(n).
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EXAMPLE
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T(4,1) = 4: (1234), (1243), (1423), (1432).
Triangle T(n,k) begins:
1;
0, 1;
0, 1, 1;
0, 2, 3, 1;
0, 4, 11, 6, 1;
0, 8, 40, 35, 10, 1;
0, 16, 148, 195, 85, 15, 1;
0, 32, 560, 1078, 665, 175, 21, 1;
0, 64, 2160, 5992, 5033, 1820, 322, 28, 1;
...
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MAPLE
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b:= proc(n) option remember; `if`(n=0, 1, add(expand(x*
b(n-j)*binomial(n-1, j-1)*ceil(2^(j-2))), j=1..n))
end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..n))(b(n)):
seq(T(n), n=0..12);
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MATHEMATICA
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b[n_] := b[n] = If[n == 0, 1, Sum[Expand[x*b[n-j]*
Binomial[n-1, j-1]*Ceiling[2^(j-2)]], {j, n}]];
T[n_] := CoefficientList[b[n], x];
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CROSSREFS
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Columns k=0-10 give: A000007, A166444, A346317, A346318, A346319, A346320, A346321, A346322, A346323, A346324, A346325.
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KEYWORD
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AUTHOR
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STATUS
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approved
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A346415
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Triangle T(n,k), n>=0, 0<=k<=n, read by rows, where column k is (1/k!) times the k-fold exponential convolution of Fibonacci numbers with themselves.
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+30
5
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1, 0, 1, 0, 1, 1, 0, 2, 3, 1, 0, 3, 11, 6, 1, 0, 5, 35, 35, 10, 1, 0, 8, 115, 180, 85, 15, 1, 0, 13, 371, 910, 630, 175, 21, 1, 0, 21, 1203, 4494, 4445, 1750, 322, 28, 1, 0, 34, 3891, 22049, 30282, 16275, 4158, 546, 36, 1, 0, 55, 12595, 107580, 202565, 144375, 49035, 8820, 870, 45, 1
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OFFSET
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0,8
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COMMENTS
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The sequence of column k>0 satisfies a linear recurrence with constant coefficients of order k+1.
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LINKS
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EXAMPLE
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Triangle T(n,k) begins:
1;
0, 1;
0, 1, 1;
0, 2, 3, 1;
0, 3, 11, 6, 1;
0, 5, 35, 35, 10, 1;
0, 8, 115, 180, 85, 15, 1;
0, 13, 371, 910, 630, 175, 21, 1;
0, 21, 1203, 4494, 4445, 1750, 322, 28, 1;
0, 34, 3891, 22049, 30282, 16275, 4158, 546, 36, 1;
0, 55, 12595, 107580, 202565, 144375, 49035, 8820, 870, 45, 1;
...
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MAPLE
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b:= proc(n) option remember; `if`(n=0, 1, add(expand(x*b(n-j)
*binomial(n-1, j-1)*(<<0|1>, <1|1>>^j)[1, 2]), j=1..n))
end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..n))(b(n)):
seq(T(n), n=0..12);
# second Maple program:
b:= proc(n, k) option remember; `if`(k=0, 0^n, `if`(k=1,
combinat[fibonacci](n), (q-> add(binomial(n, j)*
b(j, q)*b(n-j, k-q), j=0..n))(iquo(k, 2))))
end:
T:= (n, k)-> b(n, k)/k!:
seq(seq(T(n, k), k=0..n), n=0..12);
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MATHEMATICA
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b[n_, k_] := b[n, k] = If[k == 0, 0^n, If[k == 1, Fibonacci[n], With[{q = Quotient[k, 2]}, Sum[Binomial[n, j] b[j, q] b[n-j, k-q], {j, 0, n}]]]];
T[n_, k_] := b[n, k]/k!;
Table[Table[T[n, k], {k, 0, n}], {n, 0, 12}] // Flatten (* Jean-François Alcover, Nov 06 2021, after 2nd Maple program *)
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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A187247
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Triangle read by rows: T(n,k) is the number of permutations of [n] having k cycles with at most 2 alternating runs (it is assumed that the smallest element of the cycle is in the first position), 0<=k<=n.
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+30
4
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1, 0, 1, 0, 1, 1, 0, 2, 3, 1, 2, 4, 11, 6, 1, 16, 18, 40, 35, 10, 1, 104, 142, 178, 195, 85, 15, 1, 688, 1236, 1106, 1148, 665, 175, 21, 1, 5116, 10832, 9300, 7728, 5173, 1820, 322, 28, 1, 44224, 98492, 91680, 63284, 42168, 18165, 4284, 546, 36, 1, 438560, 964172, 974924, 627420, 378620, 181797, 53361, 9030, 870, 45, 1
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OFFSET
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0,8
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COMMENTS
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Sum of entries in row n is n!.
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LINKS
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FORMULA
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E.g.f.: G(t,z) = exp[(1/4)(t-1)(2z - 1 + exp(2z))]/(1-z).
The 4-variate g.f. H(u,v,w,z) (exponential with respect z), where u marks number of cycles with 1 alternating run, v marks number of cycles with 2 alternating runs, w marks the number of all cycles, and z marks the size of the permutation, is given by
H(u,v,w,z) = exp[(1/4)w((v-1)(exp(2z)+2z)+4(u-v)exp(z)+1-4u+3v)]/(1-z)^w.
We have G(t,z) = H(t,t,1,z).
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EXAMPLE
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T(3,2)=3 because we have (1)(23), (12)(3), and (13)(2).
T(4,0)=2 because we have (1423) and (1324).
T(4,1)=4 because we have (1234), (1243), (1342), and (1432).
Triangle starts:
1;
0,1;
0,1,1;
0,2,3,1;
2,4,11,6,1;
16,18,40,35,10,1;
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MAPLE
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G := exp((1/4)*(t-1)*(2*z-1+exp(2*z)))/(1-z): Gser := simplify(series(G, z = 0, 15)): for n from 0 to 13 do P[n] := sort(expand(factorial(n) * coeff(Gser, z, n))) end do: for n from 0 to 10 do seq(coeff(P[n], t, k), k = 0 .. n) end do; # yields sequence in triangular form
# second Maple program:
b:= proc(n) option remember; expand(
`if`(n=0, 1, add(b(n-j)*binomial(n-1, j-1)*
`if`(j=1, x, (j-1)!+2^(j-2)*(x-1)), j=1..n)))
end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..n))(b(n)):
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MATHEMATICA
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b[n_] := b[n] = Expand[If[n==0, 1, Sum[b[n-j]*Binomial[n-1, j-1]*If[j==1, x, (j-1)! + 2^(j-2)*(x-1)], {j, 1, n}]]];
T[n_] := Function[p, Table[Coefficient[p, x, i], {i, 0, n}]][b[n]];
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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A352363
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Triangle read by rows. The incomplete Bell transform of the swinging factorials A056040.
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+30
4
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1, 0, 1, 0, 1, 1, 0, 2, 3, 1, 0, 6, 11, 6, 1, 0, 6, 50, 35, 10, 1, 0, 30, 166, 225, 85, 15, 1, 0, 20, 756, 1246, 735, 175, 21, 1, 0, 140, 2932, 7588, 5761, 1960, 322, 28, 1, 0, 70, 11556, 45296, 46116, 20181, 4536, 546, 36, 1
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OFFSET
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0,8
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LINKS
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FORMULA
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Given a sequence s let s|n denote the initial segment s(0), s(1), ..., s(n).
(T(s))(n, k) = IncompleteBellPolynomial(n, k, s|n), where s(n) = n!/floor(n/2)!^2.
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EXAMPLE
|
Triangle starts:
[0] 1;
[1] 0, 1;
[2] 0, 1, 1;
[3] 0, 2, 3, 1;
[4] 0, 6, 11, 6, 1;
[5] 0, 6, 50, 35, 10, 1;
[6] 0, 30, 166, 225, 85, 15, 1;
[7] 0, 20, 756, 1246, 735, 175, 21, 1;
[8] 0, 140, 2932, 7588, 5761, 1960, 322, 28, 1;
[9] 0, 70, 11556, 45296, 46116, 20181, 4536, 546, 36, 1;
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MAPLE
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SwingNumber := n -> n! / iquo(n, 2)!^2:
for n from 0 to 9 do
seq(IncompleteBellB(n, k, seq(SwingNumber(j), j = 0..n)), k = 0..n) od;
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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A352366
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Triangle read by rows. The incomplete Bell transform of the Catalan numbers.
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+30
4
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1, 0, 1, 0, 1, 1, 0, 2, 3, 1, 0, 5, 11, 6, 1, 0, 14, 45, 35, 10, 1, 0, 42, 199, 210, 85, 15, 1, 0, 132, 938, 1309, 700, 175, 21, 1, 0, 429, 4675, 8498, 5789, 1890, 322, 28, 1, 0, 1430, 24489, 57455, 48762, 19929, 4410, 546, 36, 1
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OFFSET
|
0,8
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LINKS
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FORMULA
|
Given a sequence s let s|n denote the initial segment s(0), s(1), ..., s(n).
(T(s))(n, k) = IncompleteBellPolynomial(n, k, s|n) where s(n) = CatalanNumber(n).
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EXAMPLE
|
Triangle start:
[0] 1;
[1] 0, 1;
[2] 0, 1, 1;
[3] 0, 2, 3, 1;
[4] 0, 5, 11, 6, 1;
[5] 0, 14, 45, 35, 10, 1;
[6] 0, 42, 199, 210, 85, 15, 1;
[7] 0, 132, 938, 1309, 700, 175, 21, 1;
[8] 0, 429, 4675, 8498, 5789, 1890, 322, 28, 1;
[9] 0, 1430, 24489, 57455, 48762, 19929, 4410, 546, 36, 1;
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MAPLE
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CatalanNumber := n -> binomial(2*n, n)/(n + 1):
for n from 0 to 9 do
seq(IncompleteBellB(n, k, seq(CatalanNumber(j), j=0 .. n)), k = 0..n) od;
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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| |
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1, 0, 1, 0, 1, 1, 0, 2, 3, 1, 0, 5, 10, 6, 1, 0, 14, 36, 31, 10, 1, 0, 42, 137, 156, 75, 15, 1, 0, 132, 544, 787, 510, 155, 21, 1, 0, 429, 2235, 4017, 3331, 1380, 287, 28, 1, 0, 1430, 9445, 20809, 21405, 11411, 3255, 490, 36, 1, 0, 4862, 40876, 109486, 136921, 90665
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OFFSET
|
0,8
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COMMENTS
|
Production matrix of inverse is matrix with general term (-1)^(n-k+1)C(k,n-k+1).
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LINKS
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EXAMPLE
|
Triangle begins
1;
0, 1;
0, 1, 1;
0, 2, 3, 1;
0, 5, 10, 6, 1;
0, 14, 36, 31, 10, 1;
0, 42, 137, 156, 75, 15, 1;
0, 132, 544, 787, 510, 155, 21, 1;
0, 429, 2235, 4017, 3331, 1380, 287, 28, 1;
0, 1430, 9445, 20809, 21405, 11411, 3255, 490, 36, 1;
Production matrix of inverse is
0, 1;
0, -1, 1;
0, 0, -2, 1;
0, 0, 1, -3, 1;
0, 0, 0, 3, -4, 1;
0, 0, 0, -1, 6, -5, 1;
0, 0, 0, 0, -4, 10, -6, 1;
0, 0, 0, 0, 1, -10, 15, -7, 1;
0, 0, 0, 0, 0, 5, -20, 21, -8, 1;
0, 0, 0, 0, 0, -1, 15, -35, 28, -9, 1;
0, 0, 0, 0, 0, 0, -6, 35, -56, 36, -10, 1;
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KEYWORD
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AUTHOR
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STATUS
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approved
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